18.440 Midterm 2 Solutions, Fall 2009 1. (a) P {X < a} = a2 for a ∈ (0, 1) and thus FX (a) = a2 . Differentiating gives 2a a ∈ [0, 1] fX (a) = 0 otherwise (b) E[X] = 1 0 fX (x)xdx = 1 2 0 2x dx = 2/3. (c) The pair (X, Y ) is uniformly distributed on the triangle {(x, y) : x ∈ [0, 1], y ∈ [0, 1], x > y}. Thus, conditioned on X, Y is uniform on [0, X], so E[Y |X] = X/2. (d) Cov(X, Y ) = E[XY ] − E[X]E[Y ] = E[A1 A2 ] − E[X](E[1 − X]) = 1/4 − 2/9 = 1/36. 2. (a) As derived in lecture and on problem sets, the collection of two Poisson processes (the earthquake and the flood processes) may be constructed equivalently by first taking a Poisson point process of rate 2 + 3 = 5 and then independently declaring each point in the process to be an earthquake with probability 3/5 (and a flood otherwise). Thus, the number N of earthquakes before the first flood satisfies P {N = k} = (3/5)k (2/5). (b) One may recall that the sum of independent rate one exponentials is a gamma distribution with α = 2 and λ = 1: the density is thus f (t) = e−λttα−1 λα /Γ[α] = e−t t t ≥ 0 . 0 otherwise This can also be derived directly by letting X and Y denote the two exponential random variables and writing � t � t fX+Y (t) = fX (x)fY (t − x)dx = e−x e−(t−x) dx = te−t . x=0 x=0 1 (c) Cov(min{E, F, M }, M ) = Cov(min{E, F, M }, min{E, F, M }) +Cov(min{E, F, M }, M − min{E, F, M }) The memoryless property of exponentials implies that min{E, F, M } and M − min{E, F, M } are independent, so this becomes Var(min{E, F, M }). This is the variance of an exponential of rate 6, which is 1/36. 3. (a) E[X 2 Y 2 ] − E[XY ]2 = E[X 2 ]E[Y 2 ] − E[X]2 E[Y ]2 = E[X 2 ]E[Y 2 ] = 13 . 2 /2 et −1 (b) E[etX etY ] = E[etX ]E[etY ] = et t . 4. 100 60 (a) Cov(X, Y ) = Cov( 60 , X ) = 20. i=1 Xi , j=41 Xj ) = i=41 Cov(X √i i Since Var(X) = Var(Y ) = 60, we have ρ(X, Y ) = 20/ 60 · 60 = 1/3. (b) Let Z = 100 i=61 Xi . Then X and Z are independent with joint 2 2 density f (a, b) = √2π1√60 e−a /120 √2π1√40 e−b /80 . Conditioning on X + Z = x restricts us to the line a + b = x, so the conditional density for X will be (for some constant C) 2 2 /80 f (a) = Cf (a, x − a) = Ce−a /120 e−(x−a) R∞ where C is chosen so that −∞ f (a)da = 1. , 5. the ith year, then (a) If Xi is the multiplicative factor during u E[Xi ] = 2 · .4 + .5 · .6 = 1.1 and E Xi = 1.1100 ∼ 13780.6. (b) We need to get at least 50 “up” steps. Mean number is 40,√variance is 100(.6 · .4) = 24. So, very roughly, the chance is 1 − Φ(10/ 24) ∼ .96. Despite high expectation, investment usually loses money. 2 MIT OpenCourseWare http://ocw.mit.edu 18.440 Probability and Random Variables Spring 2014 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.